Question: If $x \oplus y = x^{2}+4y^{2}$ and $x \bigtriangledown y = x(y-6)$, find $1 \bigtriangledown (5 \oplus -1)$.
Explanation: First, find $5 \oplus -1$ $ 5 \oplus -1 = 5^{2}+4(-1)^{2}$ $ \hphantom{5 \oplus -1} = 29$ Now, find $1 \bigtriangledown 29$ $ 1 \bigtriangledown 29 = 29-6$ $ \hphantom{1 \bigtriangledown 29} = 23$.